derivative of this function?

1. Let the function f(x) have the property that f′(x)=x+1/x−3. If g(x)=f(x^2) find g′(x).

g'(x)=?
g'(x)=f'(x) at x^2 so, f'(x^2)?

(x+1)'(x-3)-(x+1)(x-3)'/(x-3)^2

in the end i get -4/(x-3)^2 and then I plug in x^2..
>-4/(x^4-6x^2+9)??

this seems to be wrong so could someone point out where my concept is flawed?
Thanks
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 Recognitions: Gold Member Science Advisor Staff Emeritus I have no idea why you are finding the second derivative of f. That is not at all relevant to the question. Use the chain rule: $g(x)= f(x^2)$ so $g'(x)= f'(x^2)(x^2)'$. You are given f', not f, so you do not need to do that derivative.
 Thanks so much, I don't exactly know what I was doing, either...haha I should have read the question more clearly.

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derivative of this function?

 Quote by PandaherO 1. Let the function f(x) have the property that f′(x)=x+1/x−3. If g(x)=f(x^2) find g′(x). I've tried some steps already, however my answer is still wrong.. g'(x)=? g'(x)=f'(x) at x^2 so, f'(x^2)? (x+1)'(x-3)-(x+1)(x-3)'/(x-3)^2 in the end i get -4/(x-3)^2 and then I plug in x^2.. >-4/(x^4-6x^2+9)?? this seems to be wrong so could someone point out where my concept is flawed? Thanks
Since g(x) = f(x^2), we have g'(x) = (d/dx) f(x^2), and you can write this out using the chain rule together with your formula f'(x) = x-3 + 1/x (as you have written).

RGV

 Tags f(x), find derivative, g'(x), g(x), property